JEE Advanced · Mathematics · 28. Area Under Curves
Paragraph:
Consider the functions defined implicitly by the equation \(y^3-3 y+x=0\) on various intervals in the real line. If \(x \in(-\infty,-2) \cup(2, \infty)\), the equation implicitly defines a unique real valued differentiable function \(y=f(x)\). If \(x \in(-2,2)\), the equation implicitly defines a unique real valued differentiable function \(y=g(x)\), satisfying \(g(0)=0\).Question:
The area of the region bounded by the curve \(y=f(x)\), the \(X\)-axis and the lines \(x=a\) and \(x=b\), where \(-\infty < a < b < -2\), is
- A
\(\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x+b f(b)-a f(a)\)
- B
\(-\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x+b f(b)-a f(a)\)
- C
\(\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x-b f(b)+a f(a)\)
- D
\(-\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x-b f(b)-a f(a)\)
Answer & Solution
Correct Answer
(A)
\(\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x+b f(b)-a f(a)\)
Step-by-step Solution
Detailed explanation
Required area \(=\int_a^b y d x=\int_a^b f(x) d x\)
\[
=[f(x) \cdot x]_a^b-\int_a^b f^{\prime}(x) \cdot x d x=b f(b)-a f(a)-\int_a^b f^{\prime}(x) \cdot x d x
\]
\[
=b f(b)-a f(a)+\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x
\]
As, \(\quad f^{\prime}(x)=\frac{d y}{d x}=-\frac{1}{3\left(y^2-1\right)}=-\frac{1}{3\left[(f(x))^2-1\right]}\)
\[
=[f(x) \cdot x]_a^b-\int_a^b f^{\prime}(x) \cdot x d x=b f(b)-a f(a)-\int_a^b f^{\prime}(x) \cdot x d x
\]
\[
=b f(b)-a f(a)+\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x
\]
As, \(\quad f^{\prime}(x)=\frac{d y}{d x}=-\frac{1}{3\left(y^2-1\right)}=-\frac{1}{3\left[(f(x))^2-1\right]}\)
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